# Thread: Isomorphism in normal subgroups

1. ## Isomorphism in normal subgroups

Let G = Z4 + U(4), H = <(2,3)> and K = <(2,1)>. Show that G/H is not isomorphic to G/K.

2. Originally Posted by o&apartyrock

Let $G = \mathbb{Z}_4 \oplus U(4), \ H = <(2,3)>$ and K = <(2,1)>. Show that G/H is not isomorphic to G/K.
Hint: note that both G/H and G/K have 4 elements. show that every non-identity element of G/K has order 2 and find an element of G/H which has order 4.

this is not a part of solution but the above also proves that $G/K \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2$ and $G/H \cong \mathbb{Z}_4.$

3. How is it that G/H and G/k both have 4 element? My calculations show them having 8 elements each, since Z4 + U(4) has 8 elements. Am I incorrect?

4. Originally Posted by o&apartyrock

How is it that G/H and G/k both have 4 element? My calculations show them having 8 elements each, since Z4 + U(4) has 8 elements. Am I incorrect?
H and K have 2 elements. so |G/K|=|G|/|K|=8/2 = 4. same for G/H.

5. ## LaGrange

Do you have LaGrange's Thorem yet?
$|G/H|=\frac{|G|}{|H|}$

As you noted $|G|=|\mathbb{Z}_4 \oplus U_4|=8$
Direct computation shows $|H|=|<2,3>|=|\{(2,3) , (0,1) \}|= 2$
Similarly,
$|K|=|<2,1>|=|\{(2,1) , (0,1) \}|= 2$

So you see by the above,
$|G/H|=\frac{|G|}{|H|}=\frac{8}{2}=4$
$|G/H|=\frac{|G|}{|K|}=\frac{8}{2}=4$

Follow NCA's hints to see why $H \not \cong K$

6. Ah I see. Simple mistake on my part.

,

,

### let g = z4 ⊕ u(4), h = h(2, 3)i, and k = h(2, 1)i. prove that h is isomorphic to k and that g/h is not isomorphic to g/k

Click on a term to search for related topics.